Page 283 - Physical Principles of Sedimentary Basin Analysis

8.4 The slider-block model of stick–slip motion 265

(a) x

x = 0 x = l 0

(b) x

x = 0 x = x(t) x = l + x + vt
0
s
Figure 8.7. (a) The length of the block with the spring is l 0 , when the spring is relaxed. (b) The
(rightmost) tip of the spring moves at a constant velocity v, but the block does not move until the
spring is stretched a distance x s .Timet = 0 is when the block starts to move, and x(t) is the distance
the block has moved.

where the left-hand side is the mass times the acceleration, and the right-hand side is the
sum of the forces acting on the block. The sum is the force from the spring minus the
dynamic friction. The spring force is the spring constant k times its elongation x s +vt − x,
and t = 0 is the time when the force from the spring overcomes the static friction and
the block begins to move. The spring is therefore stretched a length x s = μ s mg/k at
t = 0. The elongation of the spring at time t is the position of its tip l 0 + x s + vt
minus the position the tip would have had if it had been unstretched, l 0 + x(t) (see
Figure 8.7). The solution of equation (8.11) gives the position of the block as a function
of time. In order to solve the second-order equation we need two boundary conditions.
The first is that the block is at position x = 0at t = 0, and the second is that the
block has initially zero velocity, dx/dt = 0at t = 0. The solution for the block
position is
v
x(t) = vt + x s − x d − sin(λt) − (x s − x d ) cos(λt) (8.12)
λ
√
as showninNote 8.2, where λ = k/m and x d = μ d mg/k. The velocity of the block is
therefore
dx
= v − v cos(λt) + λ(x s − x d ) sin(λt). (8.13)
dt
The block starts out with zero velocity and it comes to rest after a time span t 1 . The time
t 1 , for the duration of the movement, is a solution of equation dx/dt = 0, which is

2 −1 λ(x s − x d )
t 1 = π − tan (8.14)
λ v
where Note 8.3 shows the details of the solution. The block is then at position

x 1 = x(t 1 ) = vt 1 + 2(x s − x d ) (8.15)